## 414/514 – Continuity

This is homework 4, due Monday, October 31, at the beginning of lecture.

Suppose that $f: {\mathbb R}\to{\mathbb R}$ and that $a\in{\mathbb R}$. We write

$\displaystyle \lim_{x\to a^-}f(x)=L$

iff for all $\epsilon>0$ there is a $\delta>0$ such that whenever $0 then $|f(x)-L|<\epsilon$. In this case, we say that $f$ converges to $L$ as $x$ approaches $a$ from the left.

Similarly, we define convergence from the right:

$\displaystyle \lim_{x\to a^+}f(x)=R$

iff for all $\epsilon>0$ there is a $\delta>0$ such that whenever $0 then $|f(x)-R|<\epsilon$. In this case, we say that $f$ converges to $R$ as $x$ approaches $a$ from the right.

In terms of these notions, we see that $\lim_{x\to a}f(x)$ exists iff both $\lim_{x\to a^-}f(x)$ and $\lim_{x\to a^+}f(x)$ exist, and are equal, in which case $\lim_{x\to a}f(x)$ equals their common value.

The function $f$ is said to have a jump discontinuity (or a type I discontinuity, or a simple discontinuity, or a discontinuity of the first kind) at $a$, iff both $\lim_{x\to a^-}f(x)$ and $\lim_{x\to a^+}f(x)$ exist, but either they are not equal, or they are equal to each other, but not to $f(a)$. Sometimes the notation $f(a-)$ is used for $\lim_{x\to a^-}f(x)$, provided that it exists and,  similarly, $f(a+)$ is used to denote $\lim_{x\to a^+}f(x)$, if it exists.

If $f$ is discontinuous at $a$, but not through a jump discontinuity, we say that $f$ has a type II discontinuity (or a discontinuity of the second kind) at $a$.

Recall that $f$ is monotone iff it is either monotonically increasing or monotonically decreasing. The first alternative means that $f(a)\le f(b)$ whenever $a\le b$. The second means that $f(a)\ge f(b)$ whenever $a\le b$.

• Suppose that $f$ is monotonically increasing. Then $f(x-)$ and $f(x+)$ exist for all $x\in{\mathbb R}$. Moreover,

$f(x-)\le f(x)\le f(x+)$,

and

$f(x+)\le f(y-)$

for all $x.

• Show that if $f$ is monotone, then it has only countably many discontinuities, and all of them are jump discontinuities.
• Let $C\subseteq {\mathbb R}$ be an arbitrary countable set. Enumerate $C$ as $\{c_i\mid i\in{\mathbb N}\}$. Given an $x\in{\mathbb R}$, let

$\displaystyle\sum_{c_n

denote the sum of all the $1/2^n$ for which $n$ satisfies that $c_n. Define a function $f:{\mathbb R}\to{\mathbb R}$ by

$\displaystyle f(x)=\sum_{c_n

for all $x$. Show that $f$ is monotonically increasing, discontinuous at all points of $C$, and continuous everywhere else.

• Give an example of a function $f$ that is discontinuous everywhere. Give an example where $f$ is continuous everywhere except at $0$, where it has a discontinuity of the second kind.

Now that we are mentioning countable sets and series, it seems like a good opportunity to introduce the following generalization of the notion of series we have been studying: Let $I$ be a set; $I$ may be finite or infinite, and it may even be uncountable. For each $i\in I$ let $c_i\ge0$ be a nonnegative real number. Define

$\displaystyle \sum_{i\in I}c_i$

as the supremum of the set $F$ consisting of all sums of the form $\sum_{i\in J}c_j$ where $J\subseteq I$ is finite. (We understand an empty sum to add to 0.)

• Show that if $I={\mathbb N}$, this definition coincides with the usual one.
• Show that if $\sum_{i\in I}c_i<+\infty$, then there is a countable set $J\subseteq I$ ($J$ may very well be infinite) such that $c_i=0$ for all $i\notin J$.

The definition of continuity in terms of $\epsilon$ and $\delta$ has the annoying property that there does not seem to be a canonical $\delta$ that one can choose at each instance. The situation is actually not too bad, as the following following result indicates: (I learned this from Ali Enayat; the result has been discovered a few times. For example, also by G. Artico and U. Marconi.)

• Show that for any continuous $f:{\mathbb R}\to{\mathbb R}$ there is a continuous function $\delta:{\mathbb R}\times(0,\infty)\to(0,\infty)$ such that for any $x\in{\mathbb R}$ and any $\epsilon>0$, if $|x-y|<\delta(x,\epsilon)$, then $|f(x)-f(y)|<\epsilon$. (One could say “For every $\epsilon$ there continuously exist a $\delta$.”)

There are several ways of approaching the above exercise. Here is a suggestion (an argument of G. de Marco), but feel free to use a different proof. First, show that

$A=\{(x,y,\epsilon)\in{\mathbb R}\times{\mathbb R}\times(0,\infty)\mid |f(x)-f(y)|<\epsilon\}$

is open. Let $A^c$ denote its complement. Consider the metric on ${\mathbb R}\times{\mathbb R}\times(0,\infty)$ given by

$d((x_1,y_1,\epsilon_1),(x_2,y_2,\epsilon_2))=|x_1-x_2|+|y_1-y_2|+|\epsilon_1-\epsilon_2|.$

(This is called the “box (or taxicab) distance.” The open sets it defines are exactly the same as the usual open sets, although the open balls are very different. You can assume this property.) Define $\delta(x,\epsilon)=d((x,x,\epsilon),A^c)$.

Enayat’s argument is longer but perhaps easier to follow, it can be found here. Whether you follow De Marco’s argument or Enayat’s make sure the write up is your own and all relevant details are verified.

(One can also show that, restricting ourselves to continuous functions $f:[0,1]\to{\mathbb R}$, the function $\delta$ can be made to depend continuously not just on $x$ and $\epsilon$, but also on $f$ (thinking of $f$ as a point in the space $C([0,1])$ with its usual sup-metric).)